A008336 a(n+1) = a(n)/n if n|a(n) else a(n)*n, a(1) = 1.

1, 1, 2, 6, 24, 120, 20, 140, 1120, 10080, 1008, 11088, 924, 12012, 858, 12870, 205920, 3500640, 194480, 3695120, 184756, 3879876, 176358, 4056234, 97349616, 2433740400, 93605400, 2527345800, 90262350, 2617608150, 87253605, 2704861755, 86555576160, 2856334013280

Offset: 1

Author

Bernardo Recamán

Keywords

• nonn
• easy
• nice
• look

Comments

The graph of log_10(a(n)+1) seems to suggest that log(a(n)) is asymptotic to C*n where C is approximately 0.8. - Daniel Forgues, Sep 18 2011

Comments from N. J. A. Sloane, Apr 14 2024: (Start)

See A370968 for the terms in increasing order with duplicates omitted.

See A337486 and A195504 for the n such that a(n+1) = a(n)/n.

Guy and Nowakowski give bounds on a(n).

Theorem: 1 is the only repeated term.

Proof: Write a(n) for A008336(n).

Suppose, seeking a contradiction, that for 1 < r < s we have a(r) = a(s).

This means that a(r)*r^e_0*(r+1)^e_1*(r+2)^e_2*...(s-1)^e_t = a(s) = a(r),

where the exponents e_* are +1 or -1. The product (P1, say) of the terms with exponent +1 must equal the product (P2, say) of the terms with exponent -1. Since r>1, we need s >= r+2.

The product P1*P2 = P1^2 of all these terms is (s-1)!/(r-1)!.

But this contradicts Erdos's theorem (Erdos 1939) that the product of two or more consecutive integers is never a square. QED.

(End)

References

• P. Erdos, On the product of consecutive integers, J. London Math. Soc., 14 (1939), 194-198.

Links

• Indranil Ghosh, Table of n, a(n) for n = 1..2732 (terms 1..1000 from T. D. Noe)
• R. K. Guy and R. Nowakowski, Unsolved Problems, Amer. Math. Monthly, vol. 102 (1995), 921-926; circa page 924.
• R. K. Guy and R. Nowakowski, Annotated extract from previous link
• Nick Hobson, Python program for this sequence
• N. J. A. Sloane, My favorite integer sequences, in Sequences and their Applications (Proceedings of SETA '98).
• Index entries for sequences related to Recamán's sequence

Crossrefs

Cf. A005132 (the original Recaman sequence).

A065422 and A260850 are variants of the present sequence.

Cf. also A195504 = Product of numbers up to n-1 used as divisors in A008336(n), n >= 2; a(1) = 1.

Cf. also A337486, A370968.

Programs

Haskell

a008336 n = a008336_list !! (n-1)
a008336_list = 1 : zipWith (/*) a008336_list [1..] where
    x /* y = if x `mod` y == 0 then x `div` y else x*y
-- Reinhard Zumkeller, Feb 22 2012, Oct 25 2010

Maple

A008336 := proc(n) option remember; if n = 1 then 1 elif A008336(n-1) mod (n-1) = 0 then A008336(n-1)/(n-1) else A008336(n-1)*(n-1); fi; end;

Mathematica

a[n_] := a[n] = If[ Divisible[ a[n-1], n-1], a[n-1]/(n-1), a[n-1]*(n-1)]; a[1] = 1; Table[a[n], {n, 1, 28}] (* Jean-François Alcover, Dec 02 2011 *)
nxt[{n_,a_}]:={n+1,If[Divisible[a,n],a/n,n*a]}; Transpose[ NestList[ nxt,{1,1},30]][[2]] (* Harvey P. Dale, May 09 2016 *)

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A008336(n):
    if n == 1: return 1
    a, b = divmod(c:=A008336(n-1),n-1)
    return c*(n-1) if b else a # Chai Wah Wu, Apr 11 2024